`(1+x)^(n)=a_(0)+a_(1)x+a_(2)x^(2) +......+a_(n)x^(n)` then
Find the sum of the series ` a_(0) +a_(2)+a_(4) +……`

Options

A. `2^(n)`
B. `2^(n-1)`
C. 2
D. `2^(n -2)`

Answer

To solve the problem, we need to find the sum of the series \( a_0 + a_2 + a_4 + \ldots \) given that \( (1+x)^n = a_0 + a_1 x + a_2 x^2 + \ldots + a_n x^n \).

### Step-by-Step Solution:

1. **Identify the Binomial Expansion**:
The expression \( (1+x)^n \) can be expanded using the binomial theorem:
\[ (1+x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k \]
Here, \( a_k = \binom{n}{k} \).

2. **Substitute \( x = 1 \)**:
Substitute \( x = 1 \) into the equation:
\[ (1+1)^n = a_0 + a_1 + a_2 + \ldots + a_n \]
This simplifies to:
\[ 2^n = a_0 + a_1 + a_2 + \ldots + a_n \tag{1} \]

3. **Substitute \( x = -1 \)**:
Now substitute \( x = -1 \):
\[ (1-1)^n = a_0 - a_1 + a_2 - a_3 + \ldots + (-1)^n a_n \]
This simplifies to:
\[ 0 = a_0 - a_1 + a_2 - a_3 + \ldots + (-1)^n a_n \tag{2} \]

4. **Add Equations (1) and (2)**:
Now, add equations (1) and (2):
\[ 2^n + 0 = (a_0 + a_1 + a_2 + \ldots + a_n) + (a_0 - a_1 + a_2 - a_3 + \ldots + (-1)^n a_n) \]
This results in:
\[ 2^n = 2(a_0 + a_2 + a_4 + \ldots) \tag{3} \]
Here, all the odd-indexed terms cancel out.

5. **Solve for the Sum of Even-Indexed Coefficients**:
From equation (3), we can isolate the sum of the even-indexed coefficients:
\[ a_0 + a_2 + a_4 + \ldots = \frac{2^n}{2} = 2^{n-1} \]

### Final Answer:
Thus, the sum of the series \( a_0 + a_2 + a_4 + \ldots \) is:
\[ \boxed{2^{n-1}} \]